unit 4: formal verification -...

Unit 4 1Chang, Huang, Li, Lin, Liu

Unit 4: Formal Verification

․Course contents⎯ Logic synthesis basics⎯ Binary-decision diagram (BDD)⎯ Verification⎯ Logic optimization ⎯ Technology mapping

․Readings⎯ Chapter 11

Unit 4 2Chang, Huang, Li, Lin, Liu

Logic Synthesis & Verification․Logic synthesis programs transform Boolean

expressions or register-transfer level (RTL) description (in Verilog/VHDL/C) into logic gate networks (netlist) in a particular library. ⎯ Three different tasks

two-level combinational synthesismultilevel combinational synthesissequential synthesis

⎯ Optimization goals: minimize area, delay, and power, etc

․Verification: Checks the equivalence of a specification and an implementation.

Unit 4 3Chang, Huang, Li, Lin, Liu

Logic Synthesis & Verification․Technology-independent optimization

⎯ Works on Boolean expression equivalent.⎯ Estimates size based on # of literals.⎯ Uses don't-cares, common factor extraction (factorization), etc.

to optimize logic.⎯ Uses simple delay models.

․Technology-dependent optimization: technology mapping/library binding⎯ Maps Boolean expressions into a particular cell library. ⎯ May perform some optimizations in addition to simple mapping.⎯ Uses more accurate delay models based on cell structures.

Unit 4 4Chang, Huang, Li, Lin, Liu

Boolean Functions

․B = {0,1}, Y = {0,1,D}․A Boolean function f: Bm → Yn

⎯ f = x1 x2 + x1 x3 + x2 x3 + x1 x2 + x2 x3 + x1 x3

․Input variables: x1, x2, …․The value of the output partitions Bm into three sets

⎯ the on-set⎯ the off-set⎯ the dc-set (don’t-care set)

Unit 4 5Chang, Huang, Li, Lin, Liu

Minterms and Cubes

․A minterm is a product of all input variables or their negations. ⎯ A minterm corresponds to a single point in Bn.

․A cube is a product of the input variables or their negations. ⎯ The fewer the number of variables in the product, the

bigger the space covered by the cube.

Unit 4 6Chang, Huang, Li, Lin, Liu

Implicant and Cover

․An implicant is a cube whose points are either in the on-set or the dc-set.

․A prime implicant is an implicant that is not included in any other implicant.

․A set of prime implicants that together cover all points in the on-set (and some or all points of the dc-set) is called a prime cover.

․A prime cover is irredundant when none of its prime implicants can be removed from the cover.

․An irredundant prime cover is minimal when the cover has the minimal number of prime implicants.

Unit 4 7Chang, Huang, Li, Lin, Liu

Cover Examples․f = x1 x3 + x2 x3 + x1 x2

․f = x1 x2 + x2 x3 + x1 x3

Unit 4 8Chang, Huang, Li, Lin, Liu

Canonical Forms

․A canonical form of a Boolean function is a uniquerepresentation of the function. ⎯ It can be used for verification purposes.

․The truth table or the sum of minterms are canonical forms⎯ They grow exponentially with the number of input variables.

․A prime irredundant cover is not a canonical form. ․Reduced ordered binary decision diagram (ROBDD):

a canonical form that is interesting from a practical point of view.

Unit 4 9Chang, Huang, Li, Lin, Liu

Logic Synthesis in Practice

․Specify the logic-level behavioral description of the circuit in some hardware-description language.

․Extract from this description the Boolean expressions related to the logic and represent them in some suitable internal form.

․Manipulate these expressions to obtain an optimized representation (two-level or multilevel).

․Perform technology mapping, a mapping from the abstract optimized representation to a netlist of cells from a library.

Unit 4 10Chang, Huang, Li, Lin, Liu

Binary-Decision Diagram (BDD) Principles․Restriction resulting in the positive and negative

cofactors of a Boolean function:

⎯ f = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3

fx1 = x2 x3 + x2 x3 + x2 x3

fx1 = x2 x3 + x2 x3 + x2 x3

․Shannon expansion (already known to Boole) states:

․A complete expansion can be obtained by successively applying Shannon expansion on all variables of a function until either of the constant functions ’0’ or ’1’ are reached.

Unit 4 11Chang, Huang, Li, Lin, Liu

Example Ordered Binary-Decision Diagram (OBDD)

․The complete Shannon expansion can be visualized as a tree (solid lines correspond to the positive cofactors and dashed lines to negative cofactors).

f = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3

Unit 4 12Chang, Huang, Li, Lin, Liu

Creating A Reduced OBDD (ROBDD)

․An OBDD is a directed tree G(V,E).․Each vertex v ∈ V is characterized by an associated

variable φ(v), a high subtree η(v) (high(v)) and a lowsubtree λ(v) (low(v)).

․Procedure to reduce an OBDD:⎯ Merge all identical leaf vertices and appropriately redirect

their incoming edges;⎯ Proceed from bottom to top, process all vertices: if two

vertices u and v are found for which φ(u) = φ(v), η(u) =η(v), and λ(u) = λ(v), merge u and v and redirect incoming edges;

⎯ For vertices v for which η(v) = λ(v), remove v and redirect its incoming edges to η(v).

Unit 4 13Chang, Huang, Li, Lin, Liu

Reduction Example

Unit 4 14Chang, Huang, Li, Lin, Liu

ROBDD Properties․The ROBDD is a canonical representation, given a

fixed ordering of the variables.․The ROBDD is a compact representation for many

Boolean functions used in practice.․Variable ordering can greatly affect the size of an

ROBDD.⎯ E.g., the parity function of k bits:

x2

2 1 2

1

k

j j

j

f x x−

=

= ⊕∏

Unit 4 15Chang, Huang, Li, Lin, Liu

A BDD Package․A BDD package refers to a software program that can

manipulate ROBDDs. It has the following properties:⎯ Interaction with BDDs takes place through an abstract

data type (functionality is independent from the internal representation used).

⎯ It supports the conversion of some external representation of a Boolean function to the internal ROBDD representation.

⎯ It can store multiple Boolean functions, sharing all vertices that can be shared.

⎯ It can create new functions by combining existing ones (e.g., h = f • g).

⎯ It can convert the internal representation back to an external one.

Unit 4 16Chang, Huang, Li, Lin, Liu

BDD Data Structures

․A unique table(implemented by a hash table) that stores all triples already processed.

․A triple (φ,η,λ) uniquely identifies an ROBDD vertex.

Unit 4 17Chang, Huang, Li, Lin, Liu

Building an ROBDD

․The procedure directly builds the compact ROBDD structure.

․A simple symbolic computation system is assumed for the derivation of the cofactors.

․π(i) gives the ith variable from the top

Unit 4 18Chang, Huang, Li, Lin, Liu

robdd_build Example

Unit 4 19Chang, Huang, Li, Lin, Liu

ROBDD Manipulation

․Separate algorithms could be designed for each separate operator on ROBDDs, such as AND, NOR, etc.

․However, the universal if-then-else operator ‘ite’ is sufficient. z = ite(f,g,h), z equals g when f is true and equals h otherwise:

․Examples:

․The ite operator is well-suited for a recursive algorithm based on ROBDDs (φ(v) = x):

Unit 4 20Chang, Huang, Li, Lin, Liu

The ite Algorithm

Unit 4 21Chang, Huang, Li, Lin, Liu

Comments on the ite Algorithm

․The algorithm processes the variables in the order used in the BDD package. ⎯ π(i) gives the ith variable from the top; π -1(x) gives the

index position of variable x from the top.․Computation of the restrictions: suppose that F is the

root vertex of the function for which Fx should be computed:

․The calculation of is done in an analogous way.․The time complexity of the algorithm is O(|F|*|G|*|H|).

Fx = η(F) if π -1(φ(F)) = i

Unit 4 22Chang, Huang, Li, Lin, Liu

ROBDD Example: Computing G from G

G = ite(G, 0, 1)

Unit 4 23Chang, Huang, Li, Lin, Liu

ROBDD Example: Computing H from F, G, G

H = F ⊕ G= ite(F, G, G)

G

Unit 4 24Chang, Huang, Li, Lin, Liu

Composition

․The composite problem is⎯ the ROBDDs of two functions f and g are known⎯ the output of g is connected to an input of f⎯ compute the ROBDD of the composed function h,

whereh = f(x1, …, xi-1, g, xi+1, …,xn).

․Using Shannon expansion, one finds that

․Now, the restrictions have to be calculated by dedicated algorithms.

Unit 4 25Chang, Huang, Li, Lin, Liu

Positive Cofactor

Unit 4 26Chang, Huang, Li, Lin, Liu

Positive Cofactor Example: Computing Fx3

Unit 4 27Chang, Huang, Li, Lin, Liu

Variable Ordering

․Reorder adjacent variables only has a local effect on the ROBDD.

Unit 4 28Chang, Huang, Li, Lin, Liu

Variable Ordering (cont’d)

․Finding the ordering that minimizes the ROBDD size for some function is intractable.⎯ The optimal ordering may change as ROBDDs are being

manipulated.․So, an ROBDD package will try to reorder the variables

at distinct moments. ⎯ It could move one variable to the top and back to the

bottom and remember the best position. It could then repeat the procedure for the other variables.

․Another ‘‘invisible’’ feature of an ROBDD package is garbage collection.

Unit 4 29Chang, Huang, Li, Lin, Liu

The Verification Problem

․The issue is to compare a specification f to an implementation g.

․They can both be represented by ROBDDs (F resp. G).․In case of a fully specified function, verification is trivial

(pointer comparison) because of the strong canonicityof the ROBDD data structure.⎯ Strong canonicity: the representations to identical

functions are the same.․If there is a dc-set, use two functions f and d. The

implementation g is correct whenis a tautology (the expression evaluates to ’1’).

Unit 4 30Chang, Huang, Li, Lin, Liu

ROBDDs and Satisfiability

․A Boolean function is satisfiable if an assignment to its variables exists for which the function becomes ‘1’

․Any Boolean function whose ROBDD is unequal to ‘0’ is satisfiable.

․Suppose that choosing a Boolean variable xi to be ‘1’costs ci. Then, the minimum-cost satisfiabilityproblem asks to minimize:

where µ(xi) = 1 when xi = ‘1’ and µ(xi) = 0 when xi = ‘0’.․Solving minimum-cost satisfiability amounts to

computing the shortest path in an ROBDD, which can be solved in linear time.⎯ Weights: w(v, η (v)) = ci, w(v, λ (v)) = 0, variable xi = φ(v).

Unit 4 31Chang, Huang, Li, Lin, Liu

Applications to Combinatorial Optimization․Zero-one integer linear programming can be

formulated as a minimum-cost satisfiability problem.․Consider the (standard form) constraint: x1 + x2 + x3 +

x4 = 3.․It can be written as:

⎯ The first 6 sums in the product: at least 3 of the 4 variables are 1.

⎯ The last sum: at least one of the variables is 0.․Many combinatorial optimization problems can also be

directly formulated in terms of the satisfiability problem.

Unit 4 32Chang, Huang, Li, Lin, Liu

Set Covering

․Given a set S = {s1, …,sm} and a set K = {K1, …,Kn} where each Kj (1 ≤ j ≤ n) is a subset of S, find a subset Γ of K such that the union of the elements Γ covers S.

․The cost of a cover is the sum of the costs cj of the elements Kj of Γ.

․Multiple cost functions are possible. E.g., cj = 1 or cj = |Kj|.

․The problem is NP-complete for most cost functions.

Unit 4 33Chang, Huang, Li, Lin, Liu

Covering Matrix

․A covering problem can be formulated as a satisfiability problem by associating variables xj with the sets Kj: (x1+ x2 + x6) • (x1 + x4 + x5 + x6) • (x2 + x3 + x5) • (x2 + x4 +x6).

․This type of covering is called unate.․A binate covering problem has an expression where

complemented variables are allowed.

• Γ = {K3,K6} is the optimalsolution when cj = |Kj|.

• K3 is redundant in Γ ={K1,K2,K3}.

Unit 4 34Chang, Huang, Li, Lin, Liu

Example Simplification Rules in Covering

s2 dominates s3K5 dominates K4

K3 is essential

Unit 4 35Chang, Huang, Li, Lin, Liu

Technology-Independent Logic Optimization

․Two-level: minimize the # of product terms.⎯

․Multi-level: minimize the #'s of literals, variables.⎯ E.g., equations are optimized using a smaller number of literals.

․Methods/CAD tools: The Quine-McCluskey method (exponential-time exact algorithm), Espresso (heuristics for two-level logic), MIS (heuristics for multi-level logic), Synopsys, etc.

Unit 4 36Chang, Huang, Li, Lin, Liu

Two-Level Logic Synthesis

․Any Boolean function can be realized in two levels: AND-OR (sum of products), NAND-NAND, etc.

․Direct implementation of two level logic using PLAs(programmable logic arrays) is not as popular as in the nMOS days.

․Classic problems, solved e.g. by the Quine-McCluskeyalgorithm.

․Popular cost function: the number of literals in the sum of products expression.

․The goal is to find a minimal irredundant prime cover.

Unit 4 37Chang, Huang, Li, Lin, Liu

Optimality in Two-Level Logic Synthesis

A local and a global minimum

Unit 4 38Chang, Huang, Li, Lin, Liu

The Quine-McCluskey Algorithm․Calculate all prime implicants (of the union of the on-set

and dc-set).․Find the minimal cover of all minterms in the on-set by

prime implicants.․Construct the covering matrix.․Simplify the covering matrix by detecting essential

columns, row and column dominance.․What is left is the cyclic core of the covering matrix.

⎯ The covering problem can then be solved by a branch-and-bound algorithm.

․Other methods do not first enumerate all prime implicants; they use an implicit representation by means of ROBDDs.

Unit 4 39Chang, Huang, Li, Lin, Liu

The Quine-McCluskey Algorithm․F(a, b, c, d) = ∑m(2, 3, 7, 9, 11, 13) + ∑d(1, 10, 15)․Step 1: Group minterms to find prime implicants by applying

xy + xy ' = x.․Step 2: Select a minimum set of prime implicants (minimum

# of literals) to implement the original function.․Exponential-time exact algorithm, huge amounts of memory!

Unit 4 40Chang, Huang, Li, Lin, Liu

Technology Mapping

․Library-based technology mapping: standard cell design.⎯ Map a function to a limited set of pre-designed cells

․Lookup table-based technology mapping: Lucent, Xilinx FPGAs, etc.⎯ Each lookup table (LUT) can implement a very large number of

functions (e.g., all functions with 4 inputs and 1 output)

․Multiplexer-based technology mapping: Actel FPGAs, etc.⎯ Logic modules are constructed with multiplexers.

Unit 4 41Chang, Huang, Li, Lin, Liu

Standard Cell Revisited

Unit 4 42Chang, Huang, Li, Lin, Liu

Pattern Graphs for an Example Library

Unit 4 43Chang, Huang, Li, Lin, Liu

Technology Mapping

․Technology Mapping: The optimization problem of finding a minimum cost covering of the subject graph by choosing from the collection of pattern graphs for all gates in the library.

․A cover is a collection of pattern graphs such that every node of the subject graph is contained in one (or more) of the pattern graphs.

․The cover is further constrained so that each input required by a pattern graph is actually an output of some other pattern graph.

Unit 4 44Chang, Huang, Li, Lin, Liu

Trivial Covering

․Mapped into 2-input NANDs and 1-input inverters.․8 2-input NAND-gates and 7 inverters for an area cost

of 23.․Best covering?

Unit 4 45Chang, Huang, Li, Lin, Liu

Optimal Tree Covering by Dynamic Programming

․If the subject directed acyclic graph (DAG) is a tree, then a polynomial-time algorithm to find the minimum cover exists.⎯ Based on dynamic programming: optimal substructure?

overlapping subproblems?․Given: subject trees (networks to be mapped), library

cells․Consider a node n of the subject tree

⎯ Recursive assumption: For all children of n, a best match which implements the node is know.

⎯ Cost of a leaf is 0.⎯ Consider each pattern tree which matches at n, compute

cost as the cost of implementing each node which the pattern requires as an input plus the cost of the pattern.

⎯ Choose the lowest-cost matching pattern to implement n.

Unit 4 46Chang, Huang, Li, Lin, Liu

Tree-Covering by Dynamic Programming․ If the subject DAG is not a tree

⎯ Partition the subject graph into forest of trees⎯ Cover each tree optimally using the dynamic programming.⎯ Overall solution is only an approximation.

․Optimality⎯ An optimal sequence of decisions has the property that

whatever the initial state and decision are, the remaining decisions must constitute an optimal decision sequence with regard to the state resulting from the first decision.

⎯ The minimum area cover for a tree T can be derived from the minimum area covers for every node below the root of T.

Unit 4 47Chang, Huang, Li, Lin, Liu

Best Covering

․A best covering with an area of 15.․Obtained by the dynamic programming approach.